Problem: On the way to school, Karen became fixated on the puzzle game on her phone!

<image>

The game is played as follows. In each level, you have a grid with n rows and m columns. Each cell originally contains the number 0.

One move consists of choosing one row or column, and adding 1 to all of the cells in that row or column.

To win the level, after all the moves, the number in the cell at the i-th row and j-th column should be equal to gi, j.

Karen is stuck on one level, and wants to know a way to beat this level using the minimum number of moves. Please, help her with this task!

Input

The first line of input contains two integers, n and m (1 ≤ n, m ≤ 100), the number of rows and the number of columns in the grid, respectively.

The next n lines each contain m integers. In particular, the j-th integer in the i-th of these rows contains gi, j (0 ≤ gi, j ≤ 500).

Output

If there is an error and it is actually not possible to beat the level, output a single integer -1.

Otherwise, on the first line, output a single integer k, the minimum number of moves necessary to beat the level.

The next k lines should each contain one of the following, describing the moves in the order they must be done:

  * row x, (1 ≤ x ≤ n) describing a move of the form "choose the x-th row". 
  * col x, (1 ≤ x ≤ m) describing a move of the form "choose the x-th column". 



If there are multiple optimal solutions, output any one of them.

Examples

Input

3 5
2 2 2 3 2
0 0 0 1 0
1 1 1 2 1


Output

4
row 1
row 1
col 4
row 3


Input

3 3
0 0 0
0 1 0
0 0 0


Output

-1


Input

3 3
1 1 1
1 1 1
1 1 1


Output

3
row 1
row 2
row 3

Note

In the first test case, Karen has a grid with 3 rows and 5 columns. She can perform the following 4 moves to beat the level:

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In the second test case, Karen has a grid with 3 rows and 3 columns. It is clear that it is impossible to beat the level; performing any move will create three 1s on the grid, but it is required to only have one 1 in the center.

In the third test case, Karen has a grid with 3 rows and 3 columns. She can perform the following 3 moves to beat the level:

<image>

Note that this is not the only solution; another solution, among others, is col 1, col 2, col 3.

Solution (in JAVA):

Solution:
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.Comparator;

public class d {
    static BufferedReader s = new BufferedReader(new InputStreamReader(System.in));
    public static void main(String[] args) throws IOException {
        StringBuilder sb = new StringBuilder();
        String[] s = s();
        int n = i(s[0]);
        int m = i(s[1]);
        int[][] a = new int[n][m];
        for (int i = 0; i < n; i++) {
            String[] s1 = s();
            for (int j = 0; j < m; j++) {
                a[i][j] = i(s1[j]);
            }
        }
        int[][] ans = new int[n][m];
        int ans1 = 0;
        if(n<m){
        for (int i = 0; i < n; i++) {
            int min = a[i][0];
            for (int j = 1; j < m; j++) {
                min = Math.min(min, a[i][j]);
            }
            for (int j = 0; j < m; j++) {
                ans[i][j] = min;
            }
            while (min-- > 0) {
                ans1++;
                sb.append("row " + (i + 1) + "\n");
            }

        }
    /*    for(int i=0;i<n;i++){
            for(int j=0;j<m;j++)
                System.out.print(ans[i][j]+" ");
            System.out.println();
        }*/
        for (int j = 0; j < m; j++) {
            int min = a[0][j] - ans[0][j];
            for (int i = 1; i < n; i++) {
                min = Math.min(min, a[i][j] - ans[i][j]);
            }
            for (int i = 0; i < n; i++) {
                ans[i][j] += min;
            }
            while (min-- > 0) {
                ans1++;
                sb.append("col " + (j + 1) + "\n");
            }
        }
    }else

    {
        for (int j = 0; j < m; j++) {
            int min = a[0][j] - ans[0][j];
            for (int i = 1; i < n; i++) {
                min = Math.min(min, a[i][j] );
            }
            for (int i = 0; i < n; i++) {
                ans[i][j] += min;
            }
            while (min-- > 0) {
                ans1++;
                sb.append("col " + (j + 1) + "\n");
            }
        }
        for (int i = 0; i < n; i++) {
            int min = a[i][0]- ans[i][0];
            for (int j = 1; j < m; j++) {
                min = Math.min(min, a[i][j]- ans[i][j]);
            }
            for (int j = 0; j < m; j++) {
                ans[i][j] += min;
            }
            while (min-- > 0) {
                ans1++;
                sb.append("row " + (i + 1) + "\n");
            }

        }
    /*    for(int i=0;i<n;i++){
            for(int j=0;j<m;j++)
                System.out.print(ans[i][j]+" ");
            System.out.println();
        }*/

    }
        int flag=0;

        for(int i=0;i<n;i++){
            for(int j=0;j<m;j++){
                if(a[i][j]!=ans[i][j]){
                    flag=1;break;
                }
            }
        }if(flag==0){
            System.out.println(ans1);
            System.out.println(sb.toString());
        }else{
            System.out.println(-1);;
        }
    }
        static String[] s() throws IOException {
        return s.readLine().trim().split("\\s+");
    }
    static int i(String ss) {
        return Integer.parseInt(ss);
    }
    static long l(String ss) {
        return Long.parseLong(ss);
    }
}
class Node1 implements Comparator<Node1>{
    int node;long cost;
    public Node1(){

    }
    public Node1(int node,long cost){
        this.node=node;this.cost=cost;
    }

    @Override
    public int compare(Node1 o1, Node1 o2) {
        if(o1.cost>o2.cost) return 1;
        if(o1.cost==o2.cost) return 0;
        return -1;
  //      return o1.cost-o2.cost;
    }
}